Đáp án:

\(\begin{array}{l}
4,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{5\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{5\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
6,\\
x = k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}\\
\cos 2x = {\cos ^2}x - {\sin ^2}x\\
2\sin x.\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)\\
4,\\
\sin x + \sin 5x = {\cos ^2}x - {\sin ^2}x\\
 \Leftrightarrow 2\sin \dfrac{{x + 5x}}{2}.\cos \dfrac{{x - 5x}}{2} = \cos 2x\\
 \Leftrightarrow 2\sin 4x.\cos \left( { - 2x} \right) = \cos 2x\\
 \Leftrightarrow 2\sin 4x.\cos 2x = \cos 2x\\
 \Leftrightarrow 2\sin 4x.\cos 2x - \cos 2x = 0\\
 \Leftrightarrow \cos 2x.\left( {2\sin 4x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 0\\
2\sin 4x - 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k\pi \\
\sin 4x = \dfrac{1}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
4x = \dfrac{\pi }{6} + k2\pi \\
4x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{5\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
5,\\
2\sin 3x.\cos x = \sin 2x + \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {3x + x} \right) + \sin \left( {3x - x} \right) = \sin 2x + \dfrac{1}{2}\\
 \Leftrightarrow \sin 4x + \sin 2x = \sin 2x + \dfrac{1}{2}\\
 \Leftrightarrow \sin 4x = \dfrac{1}{2}\\
 \Leftrightarrow \sin 4x = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{6} + k2\pi \\
4x = \pi  - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = \dfrac{\pi }{6} + k2\pi \\
4x = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{2}\\
x = \dfrac{{5\pi }}{{24}} + \dfrac{{k\pi }}{2}
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
6,\\
2{\cos ^2}x + 3\sin x = 2\\
 \Leftrightarrow 2.\left( {1 - {{\sin }^2}x} \right) + 3\sin x = 2\\
 \Leftrightarrow 2 - 2{\sin ^2}x + 3\sin x - 2 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x + 3\sin x = 0\\
 \Leftrightarrow  - \sin x.\left( {2\sin x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
2\sin x - 3 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\sin x = \dfrac{3}{2}
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \sin x = 0 \Rightarrow x = k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)